# A Simplified Version of Leo Giugiuc's Inequality from the AMM

### Problem

### Solution 1

If one of the variables is greater than $2$ there's nothing to prove as then the sum of the cubes would exceed $8!$ Otherwise, apply the AM-GM inequality:

$\displaystyle \begin{align} (abc)^2&=(abc)(2-a)(2-b)(2-c)\\ &\le\left(\frac{a+b+c+(2-a)+(2-b)+(2-c)}{6}\right)^6=1. \end{align}$

Hence, $abc\le 1.$ Earlier, Leo Giugiuc has published an inequality

$\displaystyle a^3+b^3+c^3+5abc\ge 8.$

From this, $a^3+b^3+c^3\ge 8-5abc\ge 3.$

### Solution 2

If any one of $a$, $b$, or $c$ is $\geq \sqrt[3]{3}$, then the inequality is trivially satisfied. Let us assume that this is not the case. Thus, we can make the following substitutions:

$\displaystyle \frac{a}{2-a}=e^x,~\frac{b}{2-b}=e^y,~\frac{c}{2-c}=e^z.$

The constraint is equivalent to $x+y+z=0$. Inversion gives

$a=f(x)=\frac{2e^x}{1+e^x}.$

Note, $\displaystyle \frac{d^2~[f(x)]^3}{dx^2}=\frac{24e^{3x}(3-e^x)}{(1+e^x)^5}.$ Thus, $[f(x)]^3$ is convex if $e^x\lt 3$ or if $a\lt 3/2$ (which is true because we have assumed that $a,b,c\lt \sqrt[3]{3}\lt 3/2$). Thus, applying Jensen's inequality,

$\displaystyle \begin{align} \sum_{cycl} a^3 &= \sum_{cycl} [f(x)]^3\\ &\geq 3 \left[ f\left(\frac{1}{3}\sum_{cycl} x\right)\right]^3\\ &= 3 [f(0)]^3 = 3. \end{align}$

### Acknowledgment

The problem was kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc with a comment: "Not mine. A weaker version of one of my AMM problems. But even though, what a high class problem!" Leo also commented with a solution (Solution 1). Solution 2 is by Amit Itagi.

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